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Force on the slider,

F = 1.5 x 10^-2 N

Length, (l) = 30 cm = 30/100 m = 0.3 m

Surface tension, T = F/2l = {1.5 x 10^-2}/{2 x 0.3} = 2.5 x 10^-2 Nm^-1

A factor of 2 is taken (for S = F/2l) because there are two sides of slider acted upon by the force F.

A) A and ‘R’ are correct and ‘A’ is supported by ‘R’

Advantage of geothermal energy: 

(i) It is very economical to use. 

(ii) It does not cause any pollution.

For incompressible fluids a₁v₁ =a₂v₂ =a₃v₃ =...…. where a₁, a₂, a₃,..… are the magnitudes of velocities.
For compressible fluids : a₁v₁ρ₁ = a₂v₂ρ₂ = a₃v₃ρ₃ =..…… where ρ₁, ρ₂, ρ₃,..…. are the densities of the fluids.

(a) Ocean thermal energy 

(b) No 

(c) Ocean thermal energy is used to generate electricity in an Ocean Thermal Energy Conversion power plant (OTEC power plant). A temperature difference of 20°C or more between the surface water and deeper water is needed for the operation of these plants. The warm surface water of ocean is used to boil a liquid like ammonia or chlorofluorocarbon. The high pressure of liquid vapours is used to turn the turbine of a generator and produce electricity. 

(d) 200°C

When a razor blade is made to float on water, three forces act on the blade.

(i) Weight of the blade acting vertically downwards.

(ii) Reacting on blade exerted by the liquid surface acting, vertically upwards.

(iii) Force of surface tension on circumference of the blade acting tangentially to the liquid surface.

In this case, as no portion of razor blade is immersed in water, hence Archimede's Principle is not applicable.

According to this Principle, when a body is immersed wholly or partially in a liquid, it loses its weight which is equal to the weight of the liquid displaced by the body.

Consider a capillary tube of a radius r and opened at both ends. Let it be dipped in the liquid of surface tension T (The liquid wets the glass). Liquid rises in the tube to height h as shown in Figure.

Surface tension T acts along the tangents of the spherical meniscus of the liquid at point A and B. The reaction R = T acts just opposite to the surface tension.

Resolve R = T into two components.

(i) T cosθ which acts at every point of the meniscus in the upward direction and is responsible for the rise of liquid in the capillary tube.

(ii) T sinθ acts at right angle to the length of the capillary tube. It has no effect on the rise of liquid in the tube.

The total vertical force acting on the circular meniscus is given by

F = T cosθ x 2πr   ...(i)

Due to this force liquid rises in the tube. As the liquid rises, weight of the liquid in the tube increases. Weight of the liquid acts vertically downwards. When the weight of the liquid in the tube becomes equal to the vertical force due to surface tension, liquid stops rising further in the tube.

Volume of liquid in the capillary tube is

V = Volume of cylinder of length h and radius r + (Volume of cylinder ABCD of radius r and length r) - (Volume of hemisphere of radius r)

= πr²h + (πr²r 1/2 x 4/3πr³)

= πr²h + (πr³r - 2/3 πr³) = πr³h + π/3 r³

V = πr²(h + r/3)

If ρ be the density of the liquid, then mass of the liquid raised in the capillary tube is

m = Vρ = πr²(h + r/3)ρ

Weight of the liquid raised in the capillary tube is

W = mg = πr²(h + r/3)ρg    ....(ii)

In equilibrium W = F

or, πr²(h + r/3)ρg = T cosθ x 2πr

(h + r/3) = {2T cosθ}/{rρg}

If r << h, then r/3 can be neglected as compare θ to h.

h = {2T cosθ}/{rρg}

There will be a small rise in the capillary tube if top of capillary tube is closed. Because the rise of liquid in capillary tube due to surface tension will be opposed by the downward force exerted by the compressed air above the liquid in tube. This downward force increases with increase in height of liquid column. Therefore only a small rise of liquid column is possible in a capillary tube with a closed top.

Acc. to stoke’s law:- 

F = 6 π η r v → The viscous drag force F depends on :-

F = Viscous drag

1) η = co-efficient of viscosity 

2) r = radius of spherical body 

3) V = Velocity of body

Stoke’s law states the backward dragging force (F) acting on a small spherical body of radius r, moving through a medium having coefficient of viscosity η and velocity v is given by

F = 6πηrv

viscous drag (F) depends on 

1. coefficient of viscosity of the medium η

2. velocity of the body (v) 

3. radius of the spherical body (r)

1. A person can swim in seawater more easily than in river water.

2. Ice floats on water.

3. The ship is made of steel but its interior is made hollow by giving it a concave  shape.

(c) weight of liquid

Tension = True weight - upthrust due to oil

= Vρg- Vσg = V(ρ - σ)g

= (2 x 10^-2)³ x (8920 - 820) x 10 = 8 x 10^-6 x 8100 x 10

Let the volume of the iceberg be V. The weight of the iceberg is ρ_i Vg. If x is the fraction submerged, then the volume of water displaced is xV. The buoyant force is ρ_wxVg where r w is the density of water.

ρ_i V_g = ρ_wxVg

x = ρ_i /ρ_w = 0.917

(i) People who live in glass houses should not throw stones. (suggestion)

(ii) You should wipe your feet before coming into the house, especially during the rains. ( suggestion).

(iii) You ought todo what the teacher tells you. (more emphatic advice) 

(iv) The pupils were told that they should write more neatly.(suggestion) 

(v) Sign in front of a park: You must not walk on the grass. ( compulsion) 

(vi) You should be ashamed of yourself having made such a remark. 

(vii) He left home at 9 o'clock. He should be here any minute. 

(viii) “Whatever happened to the chocolate cake?” 

“How should I know? I have just arrived.”

No, the front wheel didn't really wobble. The author himself says that it didn’t wobble. The man was just trying to prove that he knew about bicycles to create an impression.

‘It’ refers to ball-bearings.

(i) You must do your duty irrespective of consequences. 

(ii) You should study at least for an hour everyday. 

(iii) The doctor says she must sleep eight hours every night. 

(iv) You should show respect towards elders and affection towards youngsters. 

(v) You ought to exercise regularly to stay fit. 

(vi) You should take a walk every morning. 

(vii) You ought not stand on your head. 

(viii) As he has a cold, he should go to bed.

The front wheel wobbled occasionally, but it didn’t require any attention. The bicycle was in good condition and the author was pleased with it.

When the author returned from the tool shed his friend had already taken out the front wheel of the bicycle. His friend was sitting on the ground with the wheel between his legs. He was playing with the wheel while the other part of the bicycle was lying on the gravel path beside him.

The bicycle was in good condition, but the author’s friend unnecessarily tampered with it. He first took out front-wheel and I; then the gear-case. His next victim was the chain. He really had a tough time in fixing these parts back into their place. Then he lost his temper and tried bullying the thing.

The bicycle, I was glad to see, showed spirit and the subsequent proceedings degenerated s into little else than a rough and tumble fight between the friend and the machine, One moment the bicycle would be on the gravel path and he on top of it the next, the position would be reversed-he on the gravel path and the bicycle on him. Now, he would be standing flushed with victory, the bicycle firmly fixed between his legs. But his triumph would be short-lived. By a sudden, quick movement the bicycle would free itself and turning upon him, hit him sharply over the head with one of its handles.

C. To go for a long bicycle ride

1. Cheery confidence
2. Inexplicable hopefulness

1. True

2. False

3. True

4. False

5. True

6. True

7. True

8. True

D. Both ‘A’ and ‘B’

C. The writer’s companion’s fight with the bicycle.

The man was trying to show himself off as a knowledgeable person in terms of bicycle repair.This sentence was said by the man to show his confidence in himself in terms of bicycle repair.Though the narrator tried to tell him(the man) that his friend had suggested him never to try to repair a bike with problem in the gear-case as buying a new bike would be cheaper, but the man was adamant to do so, mocking the narrator's friend's abilities and understanding about bicycles. Ultimately, the man had split the gear- case into two and continued to mess with the narrator's bicycle.

Here,

τ = 18 Nm,

ω = 200 rad s^-1

As P = τω

P = 180 x 200 = 36000 watt

The short and long hands of a clock are 4 cm and 6 cm long respectively.

In an hour the hour hand completes one rotation in 12 hours therefore in 24 hours it will complete 2 rotations.

Similarly, the minute hand completes one rotation therefore in 24 hours the minute hand will complete 24 rotations.

Distance travelled by its tip in 2 days = 4(circumference of the circle with radius 4 cm)

= (4 × 2π × 4) cm = 32π cm

(As, in 2 days, the short hand will complete 4 rounds)

In 2 days, the long hand will complete 48 rounds length moved by its tip = 48(circumference of the circle with radius 6 cm)

= (48 x 2π x 6) cm

= 576 π cm

Now,

Sum of the lengths moved

= (32π + 576π)

= 608π cm

= (608 x 3.14) cm

= 1909.12 cm

Length of the hour hand is 4 cm, which describes the radius of the path inscribed by the hour hand.

Length of the minute hand is 6 cm, which describes the radius of the path inscribed by the minute hand.

The circumference of the path inscribed by the hour hand = 2 π r

C = 2 × (22/7) × 4

C = 176/7 cm

The hour hand makes 2 revolutions in one day.

Therefore distance covered by the hour hand in 2 days = (176/7) × 2 × 2

= 100.57 cm

The distance covered by the minute hand in 1 revolution = 2 π r

C = 2 × (22/7) × 6

C = 264/7

As we know, the minute hand makes 1 revolution in one hour.

In 1 day, it makes 24 revolutions.

In 2 days, it makes 2 × 24 revolutions.

The distance covered by the minute hand in 2 days = 2 × 24 × (264/7)

= 12672/7

= 1810.28 cm

The sum of the distances travelled by the hour and minute hands in 2 days = 1810.28 + 100.57

= 1910.85 cm

Length of minute hand = \(\sqrt{21}\) cm

Angle subtend by minute hand in 1 minute = \(\frac{360°}{60°}\) = 6°

Angle subtend by minute hand in 5 minute (7-7.05) = 5 × 6 = 30°

So, 

Area described by minute hand in 5 minute = \(\frac{θ}{360°}\timesπr^2\)

= \(\frac{30}{360}\times\frac{22}7\times\sqrt{21}\times\sqrt{21}\)

= \(\frac{1}{12}\times\frac{22}7\times21\)

= 5.5 cm²

Correct option is: (B) 5 × 10⁹ N/m² 

Correct option is: (A) brittle

Correct option is: (C) modulus of elasticity

Correct option is: (C) \(\frac{mg}{\pi r^2}\)

(d) may be represented in terms of L, T and x if a ≠0.

The study was conducted in the schools of Britain with a purpose. It was to find out the children of which nation actually excelled in intelligence. It is a matter of surprise for all of us that the Indian children were declared superior to the children of other countries.

एक निश्चित उद्देश्य से ब्रिटेन के स्कूल्स ऑफ ब्रिटेन में अध्ययन कराया गया था। अध्ययन के माध्यम से यह जानने की कोशिश किया गया कि किस देश का बच्चा बौद्धिकता में सबसे आगे है। यह हमारे लिए आश्चर्य का विषय है कि भारतीय बच्चों की बौद्धिक क्षमता विश्व के अन्य सभी देशों के बच्चों के बौद्धिक क्षमता से कहीं अधिक है।

Given,
R(x) = 13x² + 26x + 15

Marginal cost MR(x) = d/dx{R(x)}

= d/dx(13x² + 26X + 15)

= 26x + 26

Putting x = 15

Then MR(x) = 26 × 15 + 26

= 390 + 26

MR (7) =416

Hence,

Marginal cost = Rs. 416

Marginal revenue 
\(\frac{dR(X)}{dx} = 26x + 26\)

Marginal revenue 

= (26 x 7) + 26 = 208.

Given T.R = R(Q) = 300Q – Q² 

M.R = \(\frac{d}{dq}\)(T.R) = 300 – 2Q 

T.C = 20 + 4Q 

M.C = \(\frac{d}{dq}\)(T.C) = 4 

M.R = M.C 

⇒ 300 – 2Q = 4 

⇒ 296 = 2Q 

⇒ Q = 148.

Given T.C = -x² + 5x + 7; T.R = 12x 

profit function = p(x) = T.R. – T.C 

= 12x -(-x² + 5x + 7) = x² + 7x + 7

Given T.C = C(x) = 5x² + 2x + 3; 

M.C = \(\frac{d}{dx}\)(T.C) = 10x + 2 

A.C = \(\frac{cx}{x}\) = \(\frac{T.c}{x}\) = \(\frac{5x^2 + 2x + 3}{x}\) 

= 5x + 2 + \(\frac{3}{x}\)

Correct Option is (C) used to prevent reflected radiation to escape outside

The conical projection in Fery’s blackbody is used to prevent reflected radiation to escape outside

(C) used to prevent reflected radiation to escape outside

Correct Option is (C) 81

Stefan Boltzman law : \(\theta = \sigma AT^4\)

then,

\(\frac{\theta _1}{\theta_2} = \frac{T_1 {^4}}{T_2{^4}}\)

\(\frac{\theta_2}{\theta _1} = \frac{T_2{^4}}{T_1 {^4}}\)

\(\frac{\theta_2}{\theta _1} = \frac{3T_1{^4}}{T_1{^4}}\)

\(\theta_2 = 81 \ \theta _1 \)

Correct option is (C) 81

Correct Option is (A) \(\frac{1}{2}\)

Acoording to Wiens displacement law :

\(\lambda T = b\)

\(\lambda_1 T_1 = \lambda_2 T_2\)

\(\frac{T_1}{T_2} = \frac{0.55}{1.10}\)

\(\frac{T_1}{T_2} = \frac{1}{2}\)

Correct option is (A) \(\frac{1}2\)

Correct option is (B) doubles

Let the speed of the sailor in still water be x km/h and that of the current y km/h. 

Speed downstream = (x + y) km/h 

Speed upstream = (x – y) km/h 

As per the question 

(x + y) × 40/60 = 8

⇒ x + y = 12 ………(i) 

When the sailor goes upstream, then 

(x – y) × 1 = 8 

x – y = 8 ………(ii) 

Adding (i) and (ii), we get 

2x = 20 

⇒ x = 10 

Putting x = 10 in (i), we have 

10 + y = 12 

⇒ y = 2 

Hence, the speeds of the sailor in still water and the current are 10 km/h and 2 km/h respectively

Correct Answer is: A) Decreases

As the light enters from rarer medium to the denser medium, the speed of light decreases but the frequency remains the same.

The relation between Speed of light, wavelength and frequency is given as:

V = nλ

Where V is speed

N = frequency of light

λ = wave length

n = v/λ

In denser medium, speed decreases, therefore wavelength also decreases. To keep frequency constant.